0
$\begingroup$

If $$C\cos(\theta) = A\cos(\theta_1) + B\cos(\theta_2),$$ then how to express $\theta$ in terms of $\theta_1$ and $\theta_2$.

I started like $$\mathcal{Re}\left\{e^{j\theta}\right\} = \mathcal{Re}\left\{e^{j\theta_1}\right\} + \mathcal{Re}\left\{e^{j\theta_2}\right\},$$ but could not proceed. Please help.

Thanks.

$\endgroup$
1
  • $\begingroup$ Make a drawing in the complex plane, the result then follows $\endgroup$
    – user408856
    Commented Jun 5, 2018 at 5:21

1 Answer 1

1
$\begingroup$

$$C\cos(\theta) = A\cos(\theta_1) + B\cos(\theta_2)$$ $$\cos(\theta) = \frac{A}{C}\cos(\theta_1) + \frac{B}{C}\cos(\theta_2)$$ For given values of $\theta_1$ and $\theta_2$, if the value of $A$ changes, obviously $\theta$ changes.

Thus $\theta$ is function of $A$.

The same, $\theta$ is function of $B$ and of $C$.

This proves that $\theta$ is not function of $\theta_1$ and $\theta_2$ only.

$\theta$ is function of $\theta_1,\theta_2, A,B$ and $C$.

So, the wording of your question is ambiguous :

  • If you expect a formula giving $\theta$ as a function of $\theta_1$ and $\theta_2$ without $A,B,C$ into it, the answer is "Obviously no such a formula exists".

  • If you expect a formula giving $\theta$ as a function of $\theta_1,\theta_2, A,B,C$ the answer is evident : $$\theta=\pm\arccos\left(\frac{A}{C}\cos(\theta_1) + \frac{B}{C}\cos(\theta_2)\right)+2n\pi$$

If no particular value for $\theta_1,\theta_2, A,B,C$ , or no particular relationship between them, are specified in the wording of the question, the above general formula cannot be simplified.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .